Methods for solving the diophantine type equation: "X^2=NY^2 - 1".

diophantine.py

from sympy import sqrt, Rational

def solve_equation_brute_force(n):
    '''
    Brute Force Method:
    This method involves iterating through possible values of xx and yy until a solution is found.
    '''
    solutions = []
    for y in range(1, 10000):  # Adjust the range as needed
        x_squared = n * y * y - 1
        if x_squared < 0:
            break
        x = int(x_squared ** 0.5)
        if x * x == x_squared:
            solutions.append((x, y))
    return solutions

# Example usage:
# n = 17
# solutions = solve_equation_brute_force(n)
# print("Solutions for x^2 = {}y^2 - 1:".format(n))
# for solution in solutions:
#     print(solution)

def solve_equation_binary_search(n):
    '''
    Binary Search
    Another approach is to use binary search over the range of possible values for yy and checking if x2=ny2−1x2=ny2−1 holds true. Here's how to implement it:
    '''
    x, y = 0, 1
    while True:
        x += 1
        while x * x > n * y * y - 1:
            y += 1
        if x * x == n * y * y - 1:
            return x, y

# Example usage:
# n = 17
# x, y = solve_equation_binary_search(n)
# print(f"x = {x}, y = {y}")

from math import isqrt

def solve_equation_continued_fractions(n, limit):
    '''
    Continued Fractions
    This method utilizes continued fractions to find solutions efficiently.
    '''
    solutions = []
    for k in range(1, limit):
        m_squared = k**2 * n - 1
        if isqrt(m_squared)**2 == m_squared:
            x = isqrt(m_squared)
            y = k
            solutions.append((x, y))
    return solutions

# Example usage:
# n = 2
# limit = 100
# solutions = solve_equation_continued_fractions(n, limit)
# print("Solutions for x^2 = {}y^2 - 1 where x, y < {}: {}".format(n, limit, solutions))

def solve_equation_iterative(n, limit):
    '''
    Iterative Method
    '''
    solutions = []
    for y in range(1, limit):
        x_squared = n * y * y - 1
        if x_squared > 0 and x_squared ** 0.5 % 1 == 0:
            x = int(x_squared ** 0.5)
            solutions.append((x, y))
    return solutions

# Example usage:
# n = 2  # Example value of n
# limit = 1000  # Limit for y values
# solutions = solve_equation_iterative(n, limit)
# if solutions:
#     for x, y in solutions:
#         print(f"Solution: x = {x}, y = {y}")
# else:
#     print("No solutions found within the limit.")
import math

def pell_lucas_method(n, num_solutions):
    solutions = []
    m, d = 0, 1
    a = int(math.sqrt(n))
    a0 = a
    h, k, h_prev, k_prev = a, 1, 1, 0
    
    while len(solutions) < num_solutions:
        m = d*a - m
        d = (n - m*m) // d
        a = (a0 + m) // d
        h, h_prev = a*h + h_prev, h
        k, k_prev = a*k + k_prev, k
        x, y = h, k
        if x**2 == n * y**2 - 1:
            solutions.append((x, y))
    
    return solutions

# Example usage
# n = 2
# num_solutions = 1

# print("\nSolutions using Pell-Lucas Method:")
# print(pell_lucas_method(n, num_solutions))


def solve_equation_binary_search(n, num_pairs):
    solutions = []
    x, y = 0, 1
    while len(solutions) < num_pairs:
        x += 1
        low, high = 0, x
        while low <= high:
            mid = (low + high) // 2
            if x**5 == n * mid**2 - 1:
                solutions.append((x, mid))
                break
            elif x**5 < n * mid**2 - 1:
                high = mid - 1
            else:
                low = mid + 1
    return solutions

# Example usage:
n = 2
num_pairs = 1
pairs = solve_equation_binary_search(n, num_pairs)
for i, (x, y) in enumerate(pairs, start=1):
    print(f"Pair {i}: x = {x}, y = {y}")